Why are by-value parameters excluded from NRVO?

Question

Imagine:

S f(S a) {
  return a;
}

Why is it not allowed to alias a and the return value slot?

S s = f(t);
S s          


        

Answer 1

I feel, because the alternative is always available for the optimization:

S& f(S& a) { return a; }  // pass & return by reference
^^^  ^^^

If f() is coded as mentioned in your example, then it's perfectly alright to assume that copy is intended or side effects are expected; otherwise why not to choose the pass/return by reference ?

Suppose if NRVO applies (as you ask) then there is no difference between S f(S) and S& f(S&)!

NRVO kicks in the situations like operator +() (example) because there is no worthy alternative.

One supporting aspect, all below function have different behaviors for copying:

S& f(S& a) { return a; }  // 0 copy
S f(S& a) { return a; } // 1 copy
S f(S a) { A a1; return (...)? a : a1; }  // 2 copies

In the 3rd snippet, if the (...) is known at compile time to be false then compiler generates only 1 copy.
This means, that compiler purposefully doesn't perform optimization when a trivial alternative is available.